Let $f(x)=\begin{cases} \dfrac{2}{x^2}&\text{for }x\leq -1 \\\\ \dfrac{x+3}{\cos(x+1)}&\text{for }-1 < x < \dfrac{\pi -2}{2} \end{cases}$ Is $f$ continuous at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $f$ to be continuous at $x=-1$, we need $\lim_{x\to -1}f(x)$ and $f(-1)$ to exist and be equal. Since $-1\leq -1$, the rule that applies to $x=-1$ is $\dfrac{2}{x^2}$. So $f(-1)=\dfrac{2}{(-1)^2}=2$. Now let's analyze $\lim_{x\to -1}f(x)$. Finding $\lim_{x\to -1^{ +}}f(x)$ For $x$ -values larger than $-1$, the appropriate rule for $f(x)$ is $\dfrac{x+3}{\cos(x+1)}$. Since $\dfrac{x+3}{\cos(x+1)}$ is continuous for $-1 < x < \dfrac{\pi -2}{2}$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -1^{ +}}f(x) \\\\ &=\lim_{x\to -1^{ +}}\dfrac{x+3}{\cos(x+1)} \gray{\dfrac{x+3}{\cos(x+1)}\text{ is the rule for }x>-1} \\\\ &=\dfrac{(-1)+3}{\cos((-1)+1)} \gray{\dfrac{x+3}{\cos(x+1)}\text{ is continuous at }x=-1} \\\\ &=2 \end{aligned}$ Finding $\lim_{x\to -1^{ -}}f(x)$ For $x$ -values smaller than $-1$, the appropriate rule for $f(x)$ is $\dfrac{2}{x^2}$. Since $\dfrac{2}{x^2}$ is continuous for $ x\leq -1$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -1^{ -}}f(x) \\\\ &=\lim_{x\to -1^{ -}}\dfrac{2}{x^2} \gray{\dfrac{2}{x^2}\text{ is the rule for }x<-1} \\\\ &=\dfrac{2}{(-1)^2} \gray{\dfrac{2}{x^2}\text{ is continuous at }x=-1} \\\\ &=2 \end{aligned}$ Conclusion We found that: $\lim_{x\to -1^{ +}}f(x)=\lim_{x\to -1^{ -}}f(x)=f(-1)=2$ Since the one-sided limits are both equal to $f(-1)$, we can determine that the two-sided limit $\lim_{x\to -1}f(x)$ is also equal to $f(-1)$, and $f$ is continuous at $x=-1$.